Question

Out of 21 tickets marked with numbers 1 to 21, three tickets are drawn at random. Find the probability that the three numbers on them are in A.P.
A. $\dfrac{10}{33}$
B. $\dfrac{101}{1330}$
C. $\dfrac{99}{1330}$
D. $\dfrac{11}{133}$

Answer 1

Hint: Probability of an outcome is favorable outcomes divided by total outcomes. So, total outcomes are ${}_{3}^{21}C$and favorable outcomes are finding the three numbers from 1 to 21 having different common differences and then add the number of possibilities of picking 3 numbers with different common differences.

Complete step by step answer:
We know that,
$\text{Probability of an outcome = }\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$
Total outcomes are the number of possibilities of randomly picking 3 numbers from the tickets marked 1 to 21 are${}_{3}^{21}C$.
${}_{3}^{21}C=1330$
Favorable outcomes are the possibilities of drawing 3 numbers which are in A.P. and having different common differences which is shown below:
No. of possibilities of drawing 3 tickets which are in A.P. with a common difference of 1:
1, 2, 3; 2, 3, 4; 3, 4, 5; ………………….; 19, 20, 21
From the above, the number of possibilities of drawing 3 tickets with common difference 1 is 19.
No. of possibilities of drawing 3 tickets which are in A.P. with a common difference of 2:
1, 3, 5; 2, 4, 6; 3, 5, 7…………………….; 17, 19, 21
From the above, the number of possibilities of drawing 3 tickets with common difference 2 is 17.
Similarly we can find the common difference for 3, 4, 5…….. and the maximum common difference that can be possible is 10.
No. of possibilities of drawing 3 tickets which are in A.P. with a common difference of 10:
1, 11, 21
From the above, the number of possibilities of drawing 3 tickets with common difference 10 is 1.
So, from the above discussion we have found that number of possibilities of different common difference is:
19, 17, 15…………….. 1
Adding the above number of possibilities we get,
We can rewrite the above sequence as 1, 3, 5 ……… 19. As you can see it is in A.P. with a common difference of 2 and number of terms are 10.
So, the summation of the above sequence is:
${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
In the above expression, a = 1 and d = 2.
$\begin{align}
  & {{S}_{n}}=\dfrac{10}{2}\left( 2(1)+\left( 10-1 \right)2 \right) \\
 & \Rightarrow {{S}_{n}}=10\left( 1+9 \right) \\
 & \Rightarrow {{S}_{n}}=100 \\
\end{align}$
So, the total favorable outcomes are 100.
And total outcomes are 1330.
Hence, the probability of drawing three tickets which are in A.P. is$\dfrac{100}{1330}=\dfrac{10}{133}$.
Hence, the correct option is (a).

Note: We have found the number of terms in the sequence 1, 3, 5 ……… 19 as follows:
As the sequence is in A.P. so the first term is 1 and the common difference is 2. So, from the last term we can find the number of terms.
19 = a + (n – 1)d
In the above equation a = 1 and d = 2,
$\begin{align}
  & 19=1+\left( n-1 \right)2 \\
 & \Rightarrow 19=2n-1 \\
 & \Rightarrow n=10 \\
\end{align}$
Hence, the number of terms is equal to 10.